Question 9 Review Exercise

Solutions of Question 9 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the sum of the first $n$ terms of the series $3+7+13+21+31+\ldots$

Using method of differences to compute the sum of the given series. $$\begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=6 \\ & a_4-a_3=21-13=8 \\ & \ldots \quad \ldots \quad \ldots \\ & \ldots \quad \cdots \quad \ldots \\ & a_n-a_{n-1}=(n-1) \text { term of the series } \\ & 4,6,8, \ldots \end{align}$$ Adding column wise, we get $$\begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & =\dfrac{n-1}{2}[2 \cdot 4+2 \cdot(n-2)] \\ & =\dfrac{n-1}{2}[8+2 n-4] \\ & =\dfrac{(n-1)(2 n+4)}{2} \\ & =(n-1)(n+2) \\ \Rightarrow a_n&=n^2+n-2+a_1 \\ \Rightarrow a_n&=n^2+n-2+3 \because a_1=3 \\ \Rightarrow a_n&=n^2+n+1\end{align}$$ Taking summation of the both sides $$\begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r+\sum_{r=1}^n 1 \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2}+n \\ & =n \cdot[\dfrac{2 n^2+3 n+1+3(n+1)+6}{6}] \\ & =n \cdot[\dfrac{2 n^2+3 n+3 n+3+6}{6}] \\ & =n \cdot \dfrac{2 n^2+6 n+9}{6}\end{align}$$ Thus the sum to $n$ terms is: $$S_n=\dfrac{n(2 n^2+6 n+9)}{6}$$

Find the sum of the first $n$ terms of the series $2+5+14+41+\ldots$

Using method of differences to compute the sum of the given series. $$\begin{align} & a_2-a_1=5-2=3 \\ & a_3-a_2=14-5=9 \\ & a_4-a_3=41-14=27\end{align}$$ $a_n-a_{n-1}=(n-1)$ term of the series $3,9,27, \ldots$

Adding column wise, we get $$a_n-a_1=3+9+27+\cdots+3^n$$ Which is a geometric series with common ratio $r=3$. Therefore, $$\begin{align} a_n-a_1&=\dfrac{3(3^{n-1}-1)}{3-1} \\ & =\dfrac{3^n-3}{2}=\dfrac{1}{2} 3^n-\dfrac{3}{2} \\ \Rightarrow a_n&=\dfrac{1}{2} 3^n-\dfrac{3}{2}+3 \because a_1=3 \\ \Rightarrow a_n&=\dfrac{1}{2} 3^n+\dfrac{3}{2}\end{align}$$ Taking summation on the both sides $$\begin{align} \sum_{r=1}^n a_r&=\dfrac{1}{2} \sum_{r=1}^n 3^r+\dfrac{3}{2} \sum_{r=1}^n 1 \\ & =\dfrac{1}{2} \cdot \dfrac{3[3^n-1]}{3-1}+\dfrac{3 n}{2} \\ & =\dfrac{3^{n+1}-3}{4}+\dfrac{3 n}{2}\\ & =\dfrac{3}{4}(3^n-1)+\dfrac{3 n}{2}\end{align}$$ Thus the sum to $n$ terms is: $$S_n=\dfrac{3}{4}(3^n-1)+\dfrac{3 n}{2}$$

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